Question

A lumber company is making boards that are 2769.0 millimeters tall. If the boards are too long they must be trimmed, and if they are too short they cannot be used. A sample of 19 boards is made, and it is found that they have a mean of 2771.6 millimeters with a variance of 196.00. Is there evidence at the 0.05 level that the boards are too long and need to be trimmed? Assume the population distribution is approximately normal.

Step 1 of 5:

State the null and alternative hypotheses.

Step 2 of 5:

Find the value of the test statistic. Round your answer to three decimal places.

Step 3 of 5:

Specify if the test is one-tailed or two-tailed.

Step 4 of 5:

Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.

Step 5 of 5:

Make the decision to reject or fail to reject the null hypothesis.

Answer 1

Based on the hypothesis test, there is not enough evidence to conclude that the boards are too long and need to be trimmed.

Step-by-step explanation:

Step 1: The null hypothesis would be that the mean height of the boards is 2769.0 millimeters, while the alternative hypothesis would be that the mean height is greater than 2769.0 millimeters, indicating that the boards are too long and need to be trimmed.

Step 2: To find the value of the test statistic, we calculate the standard error of the mean, which is the square root of the variance divided by the sample size. In this case, the standard error is the square root of 196.00 divided by 19, which equals 2.783. The test statistic is then calculated as the difference between the sample mean (2771.6) and the hypothesized mean (2769.0) divided by the standard error (2.783), giving us a test statistic value of approximately 0.478.

Step 3: Since we are testing whether the boards are too long (greater than the hypothesized mean), this is a one-tailed test.

Step 4: The decision rule for rejecting the null hypothesis at the 0.05 level is to compare the test statistic value to the critical value from the t-distribution table. For a one-tailed test with 18 degrees of freedom (19 - 1), the critical value for a significance level of 0.05 is approximately 1.734. If the test statistic value is greater than 1.734, we reject the null hypothesis.

Step 5: The calculated test statistic value of 0.478 is less than the critical value of 1.734, so we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that the boards are too long and need to be trimmed.